Background: Most circuits have more than one component, called a resistor that limits the flow of charge in the circuit. A measure of this limit on charge flow is called resistance. The simplest combinations of resistors are the series and parallel connections illustrated below. The total resistance of a combination of resistors depends on both their individual values and how they are connected. In part a of the figure, resistors labeled R sub 1, R sub 2, R sub 3, and R sub 4 are connected in series along one path of a circuit. In part b of the figure, the same resistors are connected along parallel paths of a circuit. (a) A series connection of resistors. (b) A parallel connection of resistors. MAJOR FEATURES OF RESISTORS IN SERIES 1. Series resistances add: Rs = R1 + R2 + R3 + .... 2. The same current flows through each resistor in series. 3. Individual resistors in series do not get the total source voltage, but divide it.MAJOR FEATURES OF RESISTORS IN SERIES 1. Series resistances add: Rs = R1 + R2 + R3 + .... 2. The same current flows through each resistor in series. 3. Individual resistors in series do not get the total source voltage, but divide it. MAJOR FEATURES OF RESISTORS IN PARALLEL 1. Parallel resistance is found from R. = R, + R2 + R3 L. + .., and it is smaller than any individual resistance in the combination. 2. Each resistor in parallel has the same full voltage of the source applied to it. (Power distribution systems most often use parallel connections to supply the myriad devices served with the same voltage and to allow them to operate independently.) 3. Parallel resistors do not each get the total current; they divide it. Now, about the bulb brightness ... What determines the brightness of a bulb? Is it voltage? Current? The answer is BOTH. Incandescent bulbs can be viewed in a circuit as Ohmic resistors. Bulb brightness is determined by the power, P. P for a resistor is given by: P = IV where I is the current through the bulb, and V is the voltage across the bulb. Recalling that V = IR for a resistor, we can write: P = 12R = V2/R . Using this information, discuss the questions in this exercise. You must explain your reasoning in detail for credit. If the answer is not what you initially thought, find an explanation for the correct answer. In this case,where I is the current through the bulb, and V is the voltage across the bulb. Recalling that V = IR for a resistor, we can write: P = 12R = V2/R . Using this information, discuss the questions in this exercise. You must explain your reasoning in detail for credit. If the answer is not what you initially thought, find an explanation for the correct answer. In this case, provide your original reasoning along with the correct reasoning. 1. (2 pts) How does the current through a battery relate to total resistance? Based on this, rank the networks in each box according to their equivalent resistance. You can refer to the networks as A, B, and C when ranking them, and use > , , ,